For $r = a \sin 2\theta$
$\displaystyle A = 8 \times \frac{1}{2}\int_{0}^{\pi/4} (a \sin 2\theta)^2 \, d\theta$
$\displaystyle A = 4a^2 \int_{0}^{\pi/4} \sin^2 2\theta \, d\theta$
$A = 4a^2 {\displaystyle \int_{0}^{\pi/4}} \frac{1}{2}(1 - \cos 4\theta) \, d\theta$
$A = 2a^2 {\displaystyle \int_{0}^{\pi/4}} (1 - \cos 4\theta) \, d\theta$
$A = 2a^2 \left[ \theta - \dfrac{\sin 4\theta}{4} \right]_{0}^{\pi/4}$
$A = 2a^2 \left[ \left( \dfrac{\pi}{4} - \dfrac{\sin \pi}{4} \right) - \left( 0 - \dfrac{\sin 0}{4} \right) \right]$
$A = 2a^2 \left[ \left( \dfrac{\pi}{4} - 0 \right) - \left( 0 - 0 \right) \right]$
$A = \frac{1}{2}\pi a^2$ answer
For $r = a \cos 2\theta$
$\displaystyle A = 8 \times \frac{1}{2}\int_{0}^{\pi/4} (a \cos 2\theta)^2 \, d\theta$
$\displaystyle A = 4a^2 \int_{0}^{\pi/4} \cos^2 2\theta \, d\theta$
$A = 4a^2 {\displaystyle \int_{0}^{\pi/4}} \frac{1}{2}(1 + \cos 4\theta) \, d\theta$
$A = 2a^2 {\displaystyle \int_{0}^{\pi/4}} (1 + \cos 4\theta) \, d\theta$
$A = 2a^2 \left[ \theta + \dfrac{\sin 4\theta}{4} \right]_{0}^{\pi/4}$
$A = 2a^2 \left[ \left( \dfrac{\pi}{4} + \dfrac{\sin \pi}{4} \right) - \left( 0 + \dfrac{\sin 0}{4} \right) \right]$
$A = 2a^2 \left[ \left( \dfrac{\pi}{4} + 0 \right) - \left( 0 + 0 \right) \right]$
$A = \frac{1}{2}\pi a^2$ answer
